Talk:A-level Mathematics/OCR/M3/Kinematics

From the equation for velocity, and the two equations for acceleration, you can derive the equations of motion. These equations can be used to solve problems that seem very complicated at first. The equations are:


 * $$ \vec v_f = \vec v_i + \vec a t$$
 * $$ v_f^2= v_i^2+ 2 \vec a . \vec s$$
 * $$ \vec s = \vec v_i t+{\frac {1}{2}} \vec a t^2$$
 * $$ |s| = \frac {( |v_i|+ |v_f|)}{2} \times t$$

Where $$\vec a$$ is acceleration, $$ \vec s$$ is displacement, which can also be written as $$ \vec x_f - \vec x_i $$, $$ \vec x_i $$ and $$ \vec x_f $$ being the initial and final position vectors of the body respectively, $$t$$ is time, $$\vec v_i$$ is initial velocity and $$ \vec v_f$$ is final velocity. Note that these equations only work when an object is moving in a straight line and has constant acceleration.

Note that in the 3rd equation the dot product of the two vectors $$ \vec a $$ and $$ \vec b $$ is taken. The 4th equation gives the distance covered not the displacement. As a result, it's usage is not preferred.

The derivation of the above formulae, uses the basic definitions of velocity and acceleration, with the help of calculus.

By definiton,


 * 1) $$\  \vec v = d\vec x / dt $$
 * 2) $$\  \vec a=d\vec v/dt    $$
 * 3) $$\ \vec v d \vec v / d \vec x = \vec a $$

Where $$ \vec x $$ is position vector (from an axis), $$ \vec v $$ is velocity, and $$ \vec a $$ is accleration

Assuming $$ \vec a $$ to be constant, since the above formulae are defined under that condition only,

V-T Relation
$$\ \vec a dt = d\vec v $$

$$ \int_{0}^{t} \vec a\, dt = \int_{v_i}^{v_f} d \vec v $$

$$\ \vec at = \vec v_f - \vec v_i $$ or $$\ \vec v_f = \vec v_i + \vec a t $$

where $$ \vec v_f $$ is final velocity, and $$ \vec v_i $$ is initial velocity.

X - T Relation
$$\ d \vec x / d \vec t = /vec v $$

$$\ d \vec x = \vec v dt $$

$$\ d \vec x = (\vec v_i + \vec a t )dt $$

The last step was got by substituting the result from the previous derivation. Intergrating,

$$ \int_{x_i}^{x_f} dx =  \int_{0}^{t} (\vec v_i + \vec a t ), dt $$

$$ \vec x_f - \vec x_f = \vec v_i t + 1/2 \vec a t $$

V-X Relation
$$\ \vec v d \vec v / d \vec x = \vec a $$

$$\ \vec v d \vec v = \vec a d \vec x $$

Intergrating

$$\ \int_{v_i}^{v_f} \vec v, d \vec v = \int_{x_i}^{x_f} \vec a, d \vec x $$

$$\ 1/2(v_f ^2 - v_i ^2) = \vec a. (x_f - x_i) $$

$$\ v_f ^2 = v_i ^ 2 + 2 \vec a. (x_f - x_i) $$

Here $$ x_f - x_i $$ is nothing but the displacement of the body, and is substituted by the term $$ \vec s $$. So the final formula is,

$$\ v_f ^2 - v_i ^2 = 2 \vec a. \vec s $$

One very important thing that must be noted here is that the terms $$ v_f ^2 $$ and $$ v_i ^2 $$ are  not  the vector quantities, but the magnitude of the vector whole squared. Technically, it is: $$ v_f ^2 = \vec v_f. \vec v_f $$, i.e the dot product. Another thing to note is that the term $$ 2 \vec a. \vec s $$ is again the dot product of the acceleration and the displacement.